Problem: Let $f(x)=\sqrt{x}$. $f'(x)=$
The strategy We can first rewrite $f(x)$ as a rational power of $x$. Then, the derivative of $f$ can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Rewriting the radical as a rational power $\begin{aligned} f(x)&=\sqrt{x} \\\\ &=x^{^{\frac{1}{2}}} \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}f'(x) \\\\ &=\dfrac{d}{dx}\left(x^{^{\frac{1}{2}}}\right) \\\\ &=\dfrac{1}{2}x^{^{\frac{1}{2}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac{1}{2}x^{^{-\frac{1}{2}}} \end{aligned}$ In conclusion, we found that $f'(x)=\dfrac{1}{2}x^{^{-\frac{1}{2}}}$. This can also be written as $\dfrac{1}{2\sqrt x}$ (all equivalent forms are accepted).